Qualitatively, the two orders of drilling produce the same results
if both A and B succeed or if both would fail when drilled. So these
can be disregarded. By the symmetry assumed for probabilities success,
the probability of (A+,B-) is equal to that of (A-,B+).
However, the payoff is much bigger for the latter.
Only those two critical outcomes differentiate the two order strategies.
Drilling A first, and B only on A+, can produce only the inferior of
the two critical outcomes, (A+,B-), while the opposite strategy can
produce only the superior of these two outcomes, (A-,B+). Thus, it
is clear that the common-sense procedure is better.
The same argument appears in a correct quantitative analysis:
Given that the marginal probabilities of success for each of A & B
are 0.32, and the conditional probabilities of success for either
when the other is successful are 0.80, then it must be the case
that the joint probability for success of both is 0.256 and the
probabilities that either succeeds and the other fails are 0.064,
while the probability that both fail is the remainder, 0.616.
Now if one adopts the common sense strategy of drilling the better
prospect (B) first, and A only if B succeeds, the expected value is
EV = .256*(70-20) + .064*(40-20) + .680*(-10)
= 7.28
The opposite strategy, drilling A first and B only if A succeeds gives
EV = .256*(70-20) + .064*(30-20) + .680*(-10)
= 6.64
Drilling both, without regard to the first outcome gives
EV = .256*(70-20) + .064*(40-20) + .064*(30-20) + .616*(-20)
= 2.40
Given that the first drilling fails, the EV of continuing is obviously
negative for this structure of probabilities and costs.
Dave Krantz (dhk@stat.columbia.edu)